Answer :

This is a grouped frequency distribution.

To find x and y, we’ll need to find mean of the following distribution by assumed mean method and equate it to the given mean of the following distribution, 332.

And class size, c = 50.

[∵ (150 – 100) = (200 – 150) = (250 – 200) = … = (550 – 500) = 50]

This given data is in exclusive type.

So, let’s construct a table finding midpoints and stating frequencies.

So now, we have

∑f_{i}u_{i} = -130 – 3x

And ∑f_{i} = 146 + x + y = 200

Mean is given by

⇒ [given, mean = 332 and using the values from the table]

⇒

⇒

⇒ 43 × 4 = 130 + 3x

⇒ 172 = 130 + 3x

⇒ 3x = 172 – 130

⇒ 3x = 42

⇒

⇒ x = 14 …(i)

Since, there are 200 observations.

⇒ Sum of frequencies = 200 [Sum of frequencies depict total number of observation]

⇒ ∑f_{i} = 200

⇒ 146 + x + y = 200

⇒ x + y = 200 – 146

⇒ x + y = 54 …(ii)

Putting the value of x from equation (i) into equation (ii), we get

x + y = 54

⇒ 14 + y = 54

⇒ y = 54 – 14

⇒ y = 40

**Thus, the missing frequencies are x = 14 and y = 40.**

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